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3 years ago

The saucer for Roger's teacup has a radius of 3 centimeters. What is the saucer's circumference?

Mathematics

1 answer:

Fantom [35]3 years ago

7 0

C=dπ and r+r=d so 3+3=6 and 6π≈18.84 so the circumference would be 18.84

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Can you answer ASAP PLZ answer with work

luda_lava [24]

Answer:

$43.70

Step-by-step explanation:

90.50 - 4 = 87.50

87.50 ÷ 2 = $43.70

¯\_(ツ)_/¯

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3 years ago

Can you please help me

andrezito [222]

Answer:

its kinda dificult to explain (easier to show) but ill do my best

Step-by-step explanation:

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3 years ago

Adante begins to evaluate the expression 3 1/3 * 5 1/4

goldenfox [79]

For this case we must find the product of two mixed numbers:

$3 \frac {1} {3} = 3 + \frac {1} {3} = \frac {9 + 1} {3} = \frac {10} {3}\\5 \frac {1} {4} = 5 + \frac {1} {4} = \frac {20 + 1} {4} = \frac {21} {4}$

So, we have:

$\frac {10} {3} * \frac {21} {4} = \frac {10 * 21} {3 * 4} = \frac {210} {12} = \frac {105} {6} = \frac {35} {2}$

If we go to a mixed number we have:

$\frac {35} {2} = 17 \frac {1} {2}$

Answer:

$17 \frac {1} {2}$

4 0

3 years ago

ramon ate at a restaurant and gave the server a $4 tip the tip was 20% of the cost of the meal how much was the meal

Tomtit [17]

Answer:

The meal was $20

Step-by-step explanation:

20%= 1/5

1/5=4/? ---> one converts to four by multiplying it by 4 so you should do the the same with the denominator which is five.

5×4= 20

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2 years ago

The face of a clock has a circumference of 63 in. What is the area of the face of the clock?

ryzh [129]

Answer:

The area of the clock $= 315.41\ inch^{2}$

Step-by-step explanation:

We have been given the face of the clock that is $63\ in$

So that is also the circumference of the clock.

Since the clock is circular in shape.

So $2\pi(r)=63\ inch$

From here we will calculate the value of radius $(r)$ of the clock that is circular in shape.

Then $2\pi(r)=63\ inch =\frac{63}{2\pi} = 10.02\ in$

Now to find the area of the clock we will put this value of (r) in the equation of area of the circle.

Now $\pi (r)^{2}=\pi(10.02)^{2}=315.41\ in^{2}$

So the area of the face of the clock = $315.41\ in^{2}$

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3 years ago